phase transitions: kinetics - andreas dahlin › onewebmedia › tif015 ›...

2015-09-04

1

1Soft Matter Physics

Phase Transitions: Kinetics

2015-09-04

Andreas DahlinLecture 3/11Jones: 3.3-3.4

[email protected]

http://www.adahlin.com/

2015-09-04 Soft Matter Physics 2

Outline

We know from the previous lecture what phases to expect for a mixture of compounds

under given circumstances.

We also want to know how fast and by what mechanism a phase transition occurs!

• Kinetics of freezing (homogeneous or heterogeneous nucleation).

• Metastable liquid mixtures (nucleation again).

• Unstable liquid mixtures (spinodal decomposition).

Emphasis on the difference between equilibrium and kinetics!

mailto:[email protected]

http://www.adahlin.com/research/

2015-09-04

2


Repetition: Reaction Kinetics

There is an “activated state”, corresponding to an activation energy, after which the

energy change is just “downhill”.

The probability that a reaction occurs is an exponential function of the activation energy

with rate constant k usually following Arrhenius kinetics:

reaction progression

free

en

ergy

ΔG*

ΔG

Tk

Gk

B

exp

The “reaction progression”

can for instance be the size

of the crystal formed

during a freezing event!

There will be an energy cost associated with creating the interface (γSL for solid-liquid)

which scales with the area and an energy release associated with the freezing that must

scale with volume.

The energy change upon freezing a sphere (ΔHm given per mass) of radius r is then:


Nucleation in Freezing

SL

2

m

m

3

nuc π43

π4 r

T

THrrG

Clearly, this function has a

maximum for a certain r = r*.

Crystals with r < r* probably melt

again…

Crystals with r > r* probably

continue to grow…r < r*

r > r*

freezing

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Freezing Kinetics

We can find r* by taking the derivative:

The activation energy is then:

Since we also take the exponential of ΔG* to get the rate constant this becomes an

extremely temperature sensitive function!

In practice one usually gets homogenous nucleation for ΔT on the order of tens of

degrees…

TH

Trr

T

THr

r

rG

m

mSLSL

m

m2nuc 20π8π4

22

m

2

2

m

3

SL

SL

2

m

mSL

m

m

3

m

mSLnuc

3

π16

2π4

2

3

π4

TH

T

TH

T

T

TH

TH

TrGG

Often nucleation is instead initiated by a boundary (the walls of a vessel) or some kind

of impurity. The interfacial energy can then be much lower.


Heterogeneous Nucleation

Icehotel

http://www.icehotel.se

Why bother so much with the

mechanism of freezing? One

reason is that the quality in

terms of crystallinity of the

final solid is strongly affected

by the nucleation process!

For instance, grain boundaries

are formed where crystals

meet if they cannot coalesce.

2015-09-04

4

Assume we have a planar surface where nucleation occurs. This “catalyst” can be a wall

or the surface of a piece of dirt. Young’s equation (force balance) gives:

If the solid is a spherical cap its volume is by geometry:

The relevant areas are:

“catalyst” (C)

solid (S)

liquid (L)

θ


Nucleation on Surfaces

CLCSSL cos

cos2cos13

π 23

r

V

cos12π 2

SL rA

22

CS sinπrA

Sidenote: The use of Young’s

equation here is not obvious to

me since it is for liquid

droplets in equilibrium…

θ

r


Reduced Activation Energy

Activation energy can be derived as for homogenous nucleation. The energy of forming

a crystal on the surface with radius of curvature r is:

The activation energy becomes quite similar to that of homogenous nucleation

(exercise). Amazingly, we get a factor containing only the contact angle:

CLCS

22

SL

2

m

m23

nuc sinπcos1π2cos2cos13

π

rr

T

THrG

4

cos2cos12

hom

het

G

G

Now we can plot the relative

reduction in activation energy for

different contact angles.

The effect from heterogeneous

nucleation is strong for the “right”

type of surface!

θ

ΔG

*het/Δ

G*

ho

m

0 60 120 1800

0.2

0.4

0.6

0.8

1

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A perfectly clean glass is filled with perfectly clean water and put into a freezer at -5 °C.

The enthalpy of melting is ΔHm = 335 Jg-1, the interfacial tension between water and ice

0.03 Jm-2 and the density of ice 0.93 gcm-3.

(A) Show that homogenous nucleation is unlikely to occur.

(B) Show that ice is likely to form on the walls of the container.


Exercise 3.1


Exercise 3.1

(A) The activation energy for homogeneous nucleation is:

We have:

ΔHm = 3.35×105 Jkg-1

γSL = 0.03 Jm-2

ρ = 930 kgm-3

T = 268 K

Tm = 273 K

ΔT = 5 K

Inserting all values gives ΔG* = 1.38…×10-17 J. At 268 K this is equal to 3757kBT which

clearly is a very high activation energy.

22

m

2

2

m

3

SL

3

π16

TH

TG

2015-09-04

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Exercise 3.1

(B) Glass is a hydrophilic surface. We can expect a low contact angle of θ = 30° or less.

The activation energy of heterogeneous nucleation compared with homogenous

nucleation is reduced by the factor:

Inserting θ = 30° shows that ΔG*het is almost down to 1% of ΔG*hom, which means that it

is tens of kBT. Thus the nucleation can be expected to occur on “everyday timescales”.

4

cos2cos12


Crystal Branching

Why do ice crystals become

branched? The same effect

contributes to the complicated

shapes of snowflakes.

It appears that after nucleation and

some growth the solid generally

becomes “pointy” in shape.

The growth of sharp perturbations

in the forming crystal amplified!

We will just make a qualitative

argument here (Jones goes into

more details if you are interested).

Wikipedia: Ice Crystals

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Dendritic Growth

When the solid is formed, heat corresponding to ΔHm is released. This heat needs to

diffuse away for the surrounding liquid to remain undercooled!

On pointy structures, the heat flux is greater and the undercooling remains higher.

Sidenote: Faster growth releases more heat, so the argument is not obvious. The thermal

conductivities of the solid and the liquid must also come into play…

heat fluxcoldest spot


Reflections and Questions

?

2015-09-04

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In the unstable region, any small fluctuation in Φ will be amplified directly. The liquids

separate through spinodal decomposition.

But let us first look at the metastable region. Here small fluctuations are stable. Phase

separation can only occur through nucleation: A fluctuation must cause a reasonably

large volume with concentration similar to the opposite coexisting phase.


Liquid-Liquid Phase Separation

ΔFmix (Φ)

Φ0

r

Φ1 Φ2

Φ0


The Critical Size

radius (r)

free

en

ergy c

han

ge

(ΔF

nuc)

ΔFnuc(r)

critical radius r*

For describing nucleation in the metastable region, the theory is very similar to freezing.

Free energy change for spherical droplet of radius r of the coexisting phase scales with

volume. Extra energy cost associated with creating the interface (scales with area).

Again there must be a certain critical droplet size corresponding to an activation barrier.

nuc

23

nuc π43

π4rf

rrF

Here Δf is the free energy change

per volume inside the formed

nucleus. (Must be negative!)

Importantly, γnuc is now the

interfacial tension of the coexisting

compositions!

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Nucleation Activation Energy

If we take the derivative as in the treatment of freezing kinetics we get:

Thus we expect that nucleation events (with r > r*) occurs with a frequency proportional

to exp(-ΔF*/[kBT]).

It should be possible to relate Δf to ΔFsep (Jones does not but I give it a try). If we know

how many “lattice sites” there are in the nucleated droplet we can use ΔFsep:

Here v is a characteristic molecular volume (of a lattice site). Thus we connect the theory

of phase separation from an equilibrium perspective with the kinetics!

nuc

2nuc π8π4 rfrr

F

fr

nuc2

2

3

nuc

1

3

π16

fF

v

ΦΦΦFf

210sep ,,

A binary mixture follows the regular solution model with χ = 2.5.

(A) Show that the mixture is metastable at Φ = 0.2.

(B) Estimate the activation energy in of homogenous nucleation at 300 K. The molecules

have a volume of 7 nm3 and the interfacial tension between the coexisting compounds is

2×10-5 Nm-1.


Exercise 3.2

2015-09-04

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(A) First condition: Curvature of ΔFmix should be positive.

Also, Φ = 0.2 must be larger than the lower value of Φ for the coexisting phases!

Must be solved numerically. Testing a few values close to Φ = 0.2 gives Φ1 ≈ 0.15 as one

solution. Opposite coexisting phase is then at Φ2 ≈ 0.85.


Exercise 3.2

ΦΦΦΦΦΦTkF 15.21log1logBmix

5.251loglogBmix

ΦΦΦTk

Φ

F

5

1

11B2

mix

2

ΦΦTk

Φ

F

04

5B

2.0

2

mix

2

TkΦ

F

Φ

05.251loglog0mix

ΦΦΦ

Φ

F

(B) We calculate ΔFsep for formation of the coexisting compositions.

We have Φ0 = 0.2, Φ1 = 0.15 and Φ2 = 0.85. Inserting these values into ΔFmix gives

ΔFmix(Φ0) = -0.1004kBT and ΔFmix(Φ1) = ΔFmix(Φ2) = -0.104kBT.

If the molecular volume is 7 nm3 we have the free energy density as:

We can then use the formula for ΔF*:


Exercise 3.2

0mix2mix

12

101mix

12

02sep ΦFΦF

ΦΦ

ΦΦΦF

ΦΦ

ΦΦF

TkTkF BBsep 0036.01004.0104.015.085.0

15.02.0104.0

15.085.0

2.085.0

323

27B27kJm 1.23001038.1

107

0036.0

107

0036.0

Tkf

J103101.2

1102

3

π161

3

π16 20

23

35

2

3

nuc

f

F

2015-09-04

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Spinodal Decomposition

We now move to the kinetics of liquid-liquid phase separation for unstable mixtures, i.e.

the region which has negative curvature for ΔFmix(Φ).

What we want is Φ(x, y, z, t) to

describe the phase separation process

in space and time.

We can first make a qualitative

argument: If the new phases are formed

as very small units here and there will

be a lot of interfacial tension. If they

are formed as large units separated by

long distances the diffusion of material

will take a very long time.

There should be a characteristic length

for the phase separation!

optimum length

slow diffusion

high interfacial energy

vo

lum

e fr

acti

on

(Φ

)

position (x)

Fick’s law of diffusion usually says that the flux J (e.g. molecules per area and time) is

proportional to the concentration gradient:

Here D is the diffusion constant (m2s-1) for a particular molecule in a particular

environment (depends on almost anything). During spinodal decomposition we have

uphill diffusion against the concentration gradient. We are separating the mixture! How?


Uphill Diffusion

CDJ

JJ

2015-09-04

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Repetition: Chemical Potential

Strictly speaking it is the gradient in chemical potential which determines diffusive flux.

N is some measure of number of molecules. Boltzmann statistics makes it possible to

relate concentration to chemical potential:

Here μ° is the chemical potential at a standard state and C° is a standard concentration.

Note that we must work with relative concentrations with this formula, but Φ is indeed

dimensionless so:

PTVT N

G

N

F

,,

TkC

C

B

exp

ΦTk logB


Rewriting Fick’s Law

Note that the gradient of chemical potential is:

Fick’s first law in terms of chemical potential should thus be:

Note that the concentration C = Φ/v where v is the molecular (lattice site) volume.

Here comes the trick: The chemical potential must per definition also be possible to write

as a derivative of free energy density (f) with respect to concentration:

Sidenote: In our binary incompressible mixture, if we add A we must also remove B.

Strictly speaking we have an exchange chemical potential: μ = μA – μB

zyxΦzyxΦ

TkzyxΦTk ,,

,,,,log B

B

Tvk

DΦΦ

v

D

v

ΦDCDJ

B

vΦ

f

C

zyxfzyx

,,,,

2015-09-04

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Continuity Equation

We still miss the time dependence! This we get from the continuity equation.

For a gradient in direction x and a control volume dxdydz the material in per unit time is

J(x)dydz and out J(x + dx)dydz. In the absence of chemical reactions one thus has:

If we take the gradient of our flux, which is based on chemical potential, we get:

x

J

x

xxJxJ

zyx

zyxxJzyxJ

t

C

d

d

ddd

ddddd

xx + dx

C(x)

J(x)J(x + dx)

dydz

Φ

fΦ

Tvk

Dv

Φ

f

Tvk

DΦ

Tvk

DΦ

t

C

BBB

Φ

fΦ

Tk

D

t

Φ

B

The derivation in Jones looks quite different but is

essentially the same…


Model of Free Energy Density

So far things are quite logical. Let us summarize:

• Rewrite Fick’s law in terms of chemical potential.

• Define chemical potential in terms of free energy density and concentration.

• Get time dependence from continuity equation.

What we need now is basically an expression for the free energy density. One such

expression (no details here) is:

Here we have f0(Φ) which is the free energy density for a uniform mixture. The other

term with a coefficient κ introduces an energy penalty for gradients in concentration,

which essentially models interfacial energy.

So now we just insert this expression for f…

20 ΦΦff

2015-09-04

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Don’t Drink and Derive…

Looking at just one dimension we get something quite messy:

4

4

3

3

2

2

2

0

22

3

0

3

2

0

22

B

4

4

3

3

2

0

2

2

0

22

B

3

3

2

0

2

B

2

2

0

B

2

2

0

B

0

B

2

0

B

2

2

2

22

2

x

ΦΦ

x

Φ

x

Φ

x

Φ

Φ

f

x

Φ

Φ

fΦ

Φ

f

x

Φ

Tk

D

x

ΦΦ

x

Φ

x

Φ

x

Φ

Φ

f

xΦ

Φ

f

x

Φ

Tk

D

x

Φ

x

Φ

Φ

fΦ

xTk

D

x

Φ

Φ

f

xΦ

xTk

D

x

Φ

Φ

f

xΦ

xTk

D

x

Φ

Φx

Φ

Φ

f

xΦ

xTk

D

x

ΦΦf

ΦxΦ

xTk

D

t

Φ


Cahn-Hilliard

By simplifying things a bit one can get an approximate description of the problem, the

Cahn-Hilliard equation (Jones 1D version):

Here M is a kind of transport coefficient. Note that without the fourth derivative term we

have Fick’s second law with an effective diffusion constant:

Note that the effective diffusion constant becomes negative (in an unstable mixture).

The derivatives of f0 and M are obviously concentration dependent. However, in order to

solve the equation one can assume that they are constant and get:

Here Φ0 is the initial state.

4

4

2

2

2

0

2

2x

ΦΦM

x

Φ

Φ

fΦM

t

Φ

2

0

2

effΦ

fMD

tq

Φ

fMqqxΦtxΦ 2

2

0

22

0 2expcosconstant,

negative

2015-09-04

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Amplification Wavevector

The cosine factor contains a certain wavevector q representing the characteristic

lengthscale of the phase separation. The exponential factor also contains q and amplifies

certain values of it by the factor in the exponential:

The function has a maximum for some q! So Cahn-Hilliard shows that a characteristic

lengthscale emerges.

2

2

0

22 2 qΦ

fMqqA

wavevector

amp

lifi

cati

on

maximum

amplification

3

2

0

2

82 qMΦ

fMq

q

A

2/1

2

0

2

max4

1

Φ

fq

2

2

0

2

max8

3

Φ

fMA


Simulation of Phase Separation

Simulation of phase separation in 2D with Cahn Hilliard. Pattern is not entirely random!

Wikipedia: Cahn-Hilliard Equation

2015-09-04

16


Imaging Spinodal Composition

Experiments with biopolymers. During different stages of phase separation, one liquid

phase solidifies (gelation) so that the pattern becomes fixed and can be imaged.

Wassen et al.

Soft Matter 2014

from low (a) to high

(e) temperature

from slow (a) to fast

(e) “freezing”


Limits of Cahn-Hilliard

Initial phase: The Cahn-Hilliard solution shows a sinusoidal function, which means the

“borders” between the domains are as wide as the domains themselves.

For phase separation by nucleation the frequency of nucleation events determines the

initial domain sizes instead. (Cahn-Hilliard is not applicable.)

Intermediate phase: Eventually the borders become sharper as clearer interfaces between

coexisting compositions are formed. The domains are growing (coarsening). This is not

predicted by Cahn-Hilliard but must happen for full phase separation!

vo

lum

e fr

acti

on

(Φ

)

position (x)

coexisting

coexistinginitial

coarsened

2015-09-04

17


Late Stage Growth

In the late stage of growth, smaller domains tend to be destroyed while only the larger

ones grow. This is due to a diffusive flux generated by the increased local concentration

around surfaces of higher curvature, i.e. smaller particles!

position (x)

vo

lum

e fr

acti

on

(Φ

)

J

The domain size grows

proportional to t1/3 if

diffusion is rate

limiting rather than

release of material from

the particle.

Jones gives a more

detailed explanation…


Ostwald Ripening

The phenomenon where larger particles grow and smaller disappear is generally referred

to as Ostwald ripening. It occurs for emulsions but also for crystals.

Ostwald ripening explains the crunchyness of ice cream that has been in the freezer too

long. (After infinite time you get a single huge ice block separated from the cream!)

bulk (stable)

surface (less stable) pointy surface (much less stable)

2015-09-04

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Video: Supersaturated Solution


Video: Mentos + Coke

2015-09-04

19


Reflections and Questions

?

Derive the expression for the activation energy of heterogeneous nucleation in freezing

(see earlier in this lecture).

→


Exercise 2.3

4

cos2cos1

3

π162

22

m

2

2

m

3

SL

TH

TG

phase transitions: kinetics - andreas dahlin › onewebmedia › tif015 ›...

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